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I'm interested in finding the value of

  1. the integral of $\left\{\frac{1}{x}\right\}\cdot x$ (the fractional part of $\dfrac{1}{x}$ multiplied by $x$) on the interval $(a,b), a\ge 0$
  2. the integral of $\left\{\frac{1}{x}\right\}$ (fractional part of $\dfrac{1}{x}$) on the interval $(a,b), a\ge 0$

NOTE: $\left\{x \right\}= x-\left\lfloor x \right\rfloor $

Thanks

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It would be helpful if you clarified your question. When you say "the fractional part of 1/x multiplied by x", do you mean $(\frac{1}{x} - \lfloor{\frac{1}{x}}\rfloor)x$? Also, can you tell us how you've attempted this or where the problem cropped up? –  Alex Becker Apr 13 '11 at 7:49
    
it is well know formula for integral of fract(1/x) on (0,1), given by Euler-mascheroni constant. I'm arriving to a integral on (0,a) with 'a' depending on a parameter, its value is < 1. Moreover, I'm arriving to find an expression/calculate a similar integral from xfract(1/x) on (0,a), with no idea how to do it. –  user9532 Apr 13 '11 at 19:37
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2 Answers 2

A hint: Your integrands misbehave at the points $x_k$ where $1/x$ is an integer $k$. Therefore split each of the integrals (1) and (2) up in a sum of integrals over intervals $[x_{k+1},x_k]$.

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Thanks, I tried to split the interval and also the fract function integrating separately. I'm looking for (if exists) a schema similar with the calculus of the integral of fract function on (0,1). At least, having such schema will help me... (I do not know where to find the the proof for formula of the integral from fract(1/x) on (0,1) in terms of Euler-Mascheroni constant either...) –  user9532 Apr 13 '11 at 19:50
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For positive integers $n$, $$\int_{1/(n+1)}^{1/n} x \text{frac}(1/x)\ dx = \int_{1/(n+1)}^{1/n} x (1/x - n)\ dx = \frac{1}{2n(n+1)^2}$$ so $$\int_0^{1/n} x \text{frac}(1/x)\ dx = \sum_{j=n}^\infty \frac{1}{2j(j+1)^2} = \frac{1}{2n} - \frac{\Psi(1,n+1)}{2}$$ If $b > 0$ and $n = \lfloor 1/b \rfloor$, $$\int_0^b x \text{frac}(1/x)\ dx = \frac{1}{2(n+1)} - \frac{\Psi(1,n+2)}{2} + \int_{1/(n+1)}^b x (1/x - n-1)\ dx = \frac{1 - (b(n+1)-1)^2}{2(n+1)} - \frac{\Psi(1,n+2)}{2}$$ and $\int_a^b x \text{frac}(1/x)\ dx = \int_0^b x \text{frac}(1/x)\ dx - \int_0^a x \text{frac}(1/x)\ dx$.

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